Implicit in this is a choice of $k^s$ (if you want, this is a choice of geometric point, z, on Spec(k); $\pi_1(Spec(k), z)=Gal(k)$).

suppose f is a spliting of this short exact sequence , then consider the image of f,denote by Im(f) ,it is a subgroup of $\pi_1(X,x) $.Then is there any understanding of the the subgroup of $\pi_1(X,x) $generated by {Im(f)|all spliting f} ? And when will it be the full group $\pi_1(X,x)$?

1 Answer
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First a remark : you have to suppose $X$ geometrically connected to get this exact sequence.
I don't know what kind of understanding you expect, but the answer to the question will it always be the full group $\pi_1(X,x)$ is certainly no. The reason is that there are explicit examples (of proper smooth curves over a number field) when there are no sections at all. These examples due to Stix and Harari-Szamuely, are not easy to construct. Since a rational point provides a section up to conjugacy by functoriality of the étale fundamental group, such curves obviously possess no rational points. As you certainly know, the section conjecture predicts (in the case of proper smooth curves over a number field say) a bijection between rational points and conjugacy classes of sections, this may help to translate your question. But it looks rather difficult to me since it seems to imply a description of the set of all sections, a task supposed to be very arduous. For instance there are no known examples where one can show that the set of conjugacy classes of sections is finite non empty (for a proper smooth curve over a number field). The section conjecture, together with Faltings 's theorem on the finiteness of rational points of curves of genus larger than 2, would imply the finiteness of this set of conjugacy classes of sections, though. The description of the set of sections that you use in your question is thus supposed to be a very hard task.

I wish to know is there some case or some criterion that ,we can pick some spicific finite many k points of X ,thus fine many sections, like $p_1, ...,p_n$ , such that their images generate the full group
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TOMMar 4 '11 at 16:15